Date | May 2016 | Marks available | 2 | Reference code | 16M.1.hl.TZ2.5 |
Level | HL only | Paper | 1 | Time zone | TZ2 |
Command term | Find | Question number | 5 | Adapted from | N/A |
Question
A biased coin is tossed five times. The probability of obtaining a head in any one throw is \(p\).
Let \(X\) be the number of heads obtained.
Find, in terms of \(p\), an expression for \({\text{P}}(X = 4)\).
(i) Determine the value of \(p\) for which \({\text{P}}(X = 4)\) is a maximum.
(ii) For this value of \(p\), determine the expected number of heads.
Markscheme
\(X \sim {\text{B}}(5,{\text{ }}p)\) (M1)
\({\text{P}}(X = 4) = \left( {\begin{array}{*{20}{c}} 5 \\ 4 \end{array}} \right){p^4}(1 - p)\) (or equivalent) A1
[2 marks]
(i) \(\frac{{\text{d}}}{{{\text{d}}p}}(5{p^4} - 5{p^5}) = 20{p^3} - 25{p^4}\) M1A1
\(5{p^3}(4 - 5p) = 0 \Rightarrow p = \frac{4}{5}\) M1A1
Note: Do not award the final A1 if \(p = 0\) is included in the answer.
(ii) \({\text{E}}(X) = np = 5\left( {\frac{4}{5}} \right)\) (M1)
\( = 4\) A1
[6 marks]
Examiners report
This question was generally very well done and posed few problems except for the weakest candidates.
This question was generally very well done and posed few problems except for the weakest candidates.